/*
 * 
 * 
 * 
 * 
 * 
 * the License.  
 *
 *    
 *
 * 
 * 
 * 
 * 
 * 
 */

/*
 * Based on TimSort.java from the Android Open Source Project
 *
 *  Copyright (C) 2008 The Android Open Source Project
 *
 *  
 *  
 *  
 *
 *       
 *
 *  
 *  
 *  
 *  
 *  
 */

package drds.server.memory.unsafe.utils.sort;

import java.util.Comparator;

/**
 * A port of the Android TimSort class, which utilizes a
 * "stable, adaptive, iterative mergesort." See the method comment on sort() for
 * more details.
 * 
 * This has been kept in Java with the original style in order to match very
 * closely with the Android source code, and thus be easy to verify correctness.
 * The class is package private. We put a simple Scala wrapper
 * {@link drds.server.memory.unsafe.utils.sort.Sorter}, which is available to
 * The purpose of the port is to generalize the interface to the sort to accept
 * input data formats besides simple arrays where every element is sorted
 * individually. For instance, the AppendOnlyMap uses this to sort an Array with
 * alternating elements of the form [key, value, key, value]. This
 * generalization comes with minimal overhead -- see SortDataFormat for more
 * information.
 * 
 * We allow key reuse to prevent creating many key objects -- see
 * SortDataFormat.
 * 
 * @see drds.server.memory.unsafe.utils.sort.SortDataFormat
 * @see drds.server.memory.unsafe.utils.sort.Sorter
 */
class TimSort<K, Buffer> {

	/**
	 * This is the minimum sized sequence that will be merged. Shorter sequences
	 * will be lengthened by calling binarySort. If the entire array is less
	 * than this length, no merges will be performed.
	 * 
	 * This constant should be a power of two. It was 64 in Tim Peter's C
	 * implementation, but 32 was empirically determined to work better in this
	 * implementation. In the unlikely event that you set this constant to be a
	 * number that's not a power of two, you'll need to change the minRunLength
	 * computation.
	 * 
	 * If you decrease this constant, you must change the stackLen computation
	 * in the TimSort constructor, or you risk an ArrayOutOfBounds exception.
	 * See listsort.txt for a discussion of the minimum stack length required as
	 * a function of the length of the array being sorted and the minimum merge
	 * sequence length.
	 */
	private static final int MIN_MERGE = 32;

	private final SortDataFormat<K, Buffer> s;

	public TimSort(SortDataFormat<K, Buffer> sortDataFormat) {
		this.s = sortDataFormat;
	}

	/**
	 * A stable, adaptive, iterative mergesort that requires far fewer than n
	 * lg(n) comparisons when running on partially sorted arrays, while offering
	 * performance comparable to a traditional mergesort when run on random
	 * arrays. Like all proper mergesorts, this sort is stable and runs O(n log
	 * n) time (worst case). In the worst case, this sort requires temporary
	 * storage space for n/2 object references; in the best case, it requires
	 * only a small constant amount of space.
	 * 
	 * This implementation was adapted from Tim Peters's list sort for Python,
	 * which is described in detail here:
	 * 
	 * http://svn.python.org/projects/python/trunk/Objects/listsort.txt
	 * 
	 * Tim's C code may be found here:
	 * 
	 * http://svn.python.org/projects/python/trunk/Objects/listobject.c
	 * 
	 * The underlying techniques are described in this paper (and may have even
	 * earlier origins):
	 * 
	 * "Optimistic Sorting and Information Theoretic Complexity" Peter McIlroy
	 * SODA (Fourth Annual ACM-SIAM Symposium on Discrete Algorithms), pp
	 * 467-474, Austin, Texas, 25-27 January 1993.
	 * 
	 * While the API to this class consists solely of static methods, it is
	 * (privately) instantiable; a TimSort instance holds the state of an
	 * ongoing sort, assuming the input array is large enough to warrant the
	 * full-blown TimSort. Small arrays are sorted in place, using a binary
	 * insertion sort.
	 * 
	 * @author Josh Bloch
	 */
	public void sort(Buffer a, int lo, int hi, Comparator<? super K> c) {
		assert c != null;

		int nRemaining = hi - lo;
		if (nRemaining < 2)
			return; // Arrays of size 0 and 1 are always sorted

		// If array is small, do a "mini-TimSort" with no merges
		if (nRemaining < MIN_MERGE) {
			int initRunLen = countRunAndMakeAscending(a, lo, hi, c);
			binarySort(a, lo, hi, lo + initRunLen, c);
			return;
		}

		/**
		 * March over the array once, left to right, finding natural runs,
		 * extending short natural runs to minRun elements, and merging runs to
		 * maintain stack invariant.
		 */
		SortState sortState = new SortState(a, c, hi - lo);
		int minRun = minRunLength(nRemaining);
		do {
			// Identify next run
			int runLen = countRunAndMakeAscending(a, lo, hi, c);

			// If run is short, extend to min(minRun, nRemaining)
			if (runLen < minRun) {
				int force = nRemaining <= minRun ? nRemaining : minRun;
				binarySort(a, lo, lo + force, lo + runLen, c);
				runLen = force;
			}

			// Push run onto pending-run stack, and maybe merge
			sortState.pushRun(lo, runLen);
			sortState.mergeCollapse();

			// Advance to find next run
			lo += runLen;
			nRemaining -= runLen;
		} while (nRemaining != 0);

		// Merge all remaining runs to complete sort
		assert lo == hi;
		sortState.mergeForceCollapse();
		assert sortState.stackSize == 1;
	}

	/**
	 * Sorts the specified portion of the specified array using a binary
	 * insertion sort. This is the best method for sorting small numbers of
	 * elements. It requires O(n log n) compares, but O(n^2) data movement
	 * (worst case).
	 * 
	 * If the initial part of the specified range is already sorted, this method
	 * can take advantage of it: the method assumes that the elements from index
	 * {@code lo}, inclusive, to {@code start}, exclusive are already sorted.
	 * 
	 * @param a
	 *            the array in which a range is to be sorted
	 * @param lo
	 *            the index of the first element in the range to be sorted
	 * @param hi
	 *            the index after the last element in the range to be sorted
	 * @param start
	 *            the index of the first element in the range that is not
	 *            already known to be sorted ({@code lo <= start <= hi})
	 * @param c
	 *            comparator to used for the sort
	 */
	@SuppressWarnings("fallthrough")
	private void binarySort(Buffer a, int lo, int hi, int start, Comparator<? super K> c) {
		assert lo <= start && start <= hi;
		if (start == lo)
			start++;

		K key0 = s.newKey();
		K key1 = s.newKey();

		Buffer pivotStore = s.allocate(1);
		for (; start < hi; start++) {
			s.copyElement(a, start, pivotStore, 0);
			K pivot = s.getKey(pivotStore, 0, key0);

			// Set left (and right) to the index where a[start] (pivot) belongs
			int left = lo;
			int right = start;
			assert left <= right;
			/*
			 * Invariants: pivot >= all in [lo, left). pivot < all in [right,
			 * start).
			 */
			while (left < right) {
				int mid = (left + right) >>> 1;
				if (c.compare(pivot, s.getKey(a, mid, key1)) < 0)
					right = mid;
				else
					left = mid + 1;
			}
			assert left == right;

			/*
			 * The invariants still hold: pivot >= all in [lo, left) and pivot <
			 * all in [left, start), so pivot belongs at left. Note that if
			 * there are elements equal to pivot, left points to the first slot
			 * after them -- that's why this sort is stable. Slide elements over
			 * to make room for pivot.
			 */
			int n = start - left; // The number of elements to move
			// Switch is just an optimization for arraycopy in default case
			switch (n) {
			case 2:
				s.copyElement(a, left + 1, a, left + 2);
			case 1:
				s.copyElement(a, left, a, left + 1);
				break;
			default:
				s.copyRange(a, left, a, left + 1, n);
			}
			s.copyElement(pivotStore, 0, a, left);
		}
	}

	/**
	 * Returns the length of the run beginning at the specified position in the
	 * specified array and reverses the run if it is descending (ensuring that
	 * the run will always be ascending when the method returns).
	 * 
	 * A run is the longest ascending sequence with:
	 * 
	 * a[lo] <= a[lo + 1] <= a[lo + 2] <= ...
	 * 
	 * or the longest descending sequence with:
	 * 
	 * a[lo] > a[lo + 1] > a[lo + 2] > ...
	 * 
	 * For its intended use in a stable mergesort, the strictness of the
	 * definition of "descending" is needed so that the call can safely reverse
	 * a descending sequence without violating stability.
	 * 
	 * @param a
	 *            the array in which a run is to be counted and possibly
	 *            reversed
	 * @param lo
	 *            index of the first element in the run
	 * @param hi
	 *            index after the last element that may be contained in the run.
	 *            It is required that {@code lo < hi}.
	 * @param c
	 *            the comparator to used for the sort
	 * @return the length of the run beginning at the specified position in the
	 *         specified array
	 */
	private int countRunAndMakeAscending(Buffer a, int lo, int hi, Comparator<? super K> c) {
		assert lo < hi;
		int runHi = lo + 1;
		if (runHi == hi)
			return 1;

		K key0 = s.newKey();
		K key1 = s.newKey();

		// Find end of run, and reverse range if descending
		if (c.compare(s.getKey(a, runHi++, key0), s.getKey(a, lo, key1)) < 0) { // Descending
			while (runHi < hi && c.compare(s.getKey(a, runHi, key0), s.getKey(a, runHi - 1, key1)) < 0)
				runHi++;
			reverseRange(a, lo, runHi);
		} else { // Ascending
			while (runHi < hi && c.compare(s.getKey(a, runHi, key0), s.getKey(a, runHi - 1, key1)) >= 0)
				runHi++;
		}

		return runHi - lo;
	}

	/**
	 * Reverse the specified range of the specified array.
	 * 
	 * @param a
	 *            the array in which a range is to be reversed
	 * @param lo
	 *            the index of the first element in the range to be reversed
	 * @param hi
	 *            the index after the last element in the range to be reversed
	 */
	private void reverseRange(Buffer a, int lo, int hi) {
		hi--;
		while (lo < hi) {
			s.swap(a, lo, hi);
			lo++;
			hi--;
		}
	}

	/**
	 * Returns the minimum acceptable run length for an array of the specified
	 * length. Natural runs shorter than this will be extended with
	 * {@link #binarySort}.
	 * 
	 * Roughly speaking, the computation is:
	 * 
	 * If n < MIN_MERGE, return n (it's too small to bother with fancy stuff).
	 * Else if n is an exact power of 2, return MIN_MERGE/2. Else return an int
	 * k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k is close to, but strictly
	 * less than, an exact power of 2.
	 * 
	 * For the rationale, see listsort.txt.
	 * 
	 * @param n
	 *            the length of the array to be sorted
	 * @return the length of the minimum run to be merged
	 */
	private int minRunLength(int n) {
		assert n >= 0;
		int r = 0; // Becomes 1 if any 1 bits are shifted off
		while (n >= MIN_MERGE) {
			r |= (n & 1);
			n >>= 1;
		}
		return n + r;
	}

	private class SortState {

		/**
		 * The Buffer being sorted.
		 */
		private final Buffer a;

		/**
		 * Length of the sort Buffer.
		 */
		private final int aLength;

		/**
		 * The comparator for this sort.
		 */
		private final Comparator<? super K> c;

		/**
		 * When we get into galloping mode, we stay there until both runs win
		 * less often than MIN_GALLOP consecutive times.
		 */
		private static final int MIN_GALLOP = 7;

		/**
		 * This controls when we get *into* galloping mode. It is initialized to
		 * MIN_GALLOP. The mergeLo and mergeHi methods nudge it higher for
		 * random data, and lower for highly structured data.
		 */
		private int minGallop = MIN_GALLOP;

		/**
		 * Maximum initial size of tmp array, which is used for merging. The
		 * array can grow to accommodate demand.
		 * 
		 * Unlike Tim's original C version, we do not allocate this much storage
		 * when sorting smaller arrays. This change was required for
		 * performance.
		 */
		private static final int INITIAL_TMP_STORAGE_LENGTH = 256;

		/**
		 * Temp storage for merges.
		 */
		private Buffer tmp; // Actual runtime type will be Object[], regardless
		// of T

		/**
		 * Length of the temp storage.
		 */
		private int tmpLength = 0;

		/**
		 * A stack of pending runs yet to be merged. Run i starts at address
		 * base[i] and extends for len[i] elements. It's always true (so long as
		 * the indices are in bounds) that:
		 * 
		 * runBase[i] + runLen[i] == runBase[i + 1]
		 * 
		 * so we could cut the storage for this, but it's a minor amount, and
		 * keeping all the info explicit simplifies the code.
		 */
		private int stackSize = 0; // Number of pending runs on stack
		private final int[] runBase;
		private final int[] runLen;

		/**
		 * Creates a TimSort instance to maintain the state of an ongoing sort.
		 * 
		 * @param a
		 *            the array to be sorted
		 * @param c
		 *            the comparator to determine the order of the sort
		 */
		private SortState(Buffer a, Comparator<? super K> c, int len) {
			this.aLength = len;
			this.a = a;
			this.c = c;

			// Allocate temp storage (which may be increased later if necessary)
			tmpLength = len < 2 * INITIAL_TMP_STORAGE_LENGTH ? len >>> 1 : INITIAL_TMP_STORAGE_LENGTH;
			tmp = s.allocate(tmpLength);

			/*
			 * Allocate runs-to-be-merged stack (which cannot be expanded). The
			 * stack length requirements are described in listsort.txt. The C
			 * version always uses the same stack length (85), but this was
			 * measured to be too expensive when sorting "mid-sized" arrays
			 * (e.g., 100 elements) in Java. Therefore, we use smaller (but
			 * sufficiently large) stack lengths for smaller arrays. The
			 * "magic numbers" in the computation below must be changed if
			 * MIN_MERGE is decreased. See the MIN_MERGE declaration above for
			 * more information.
			 */
			int stackLen = (len < 120 ? 5 : len < 1542 ? 10 : len < 119151 ? 19 : 40);
			runBase = new int[stackLen];
			runLen = new int[stackLen];
		}

		/**
		 * Pushes the specified run onto the pending-run stack.
		 * 
		 * @param runBase
		 *            index of the first element in the run
		 * @param runLen
		 *            the number of elements in the run
		 */
		private void pushRun(int runBase, int runLen) {
			this.runBase[stackSize] = runBase;
			this.runLen[stackSize] = runLen;
			stackSize++;
		}

		/**
		 * Examines the stack of runs waiting to be merged and merges adjacent
		 * runs until the stack invariants are reestablished:
		 * 
		 * 1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1] 2. runLen[i - 2] >
		 * runLen[i - 1]
		 * 
		 * This method is called each time a new run is pushed onto the stack,
		 * so the invariants are guaranteed to hold for i < stackSize upon entry
		 * to the method.
		 */
		private void mergeCollapse() {
			while (stackSize > 1) {
				int n = stackSize - 2;
				if ((n >= 1 && runLen[n - 1] <= runLen[n] + runLen[n + 1]) || (n >= 2 && runLen[n - 2] <= runLen[n] + runLen[n - 1])) {
					if (runLen[n - 1] < runLen[n + 1])
						n--;
				} else if (runLen[n] > runLen[n + 1]) {
					break; // Invariant is established
				}
				mergeAt(n);
			}
		}

		/**
		 * Merges all runs on the stack until only one remains. This method is
		 * called once, to complete the sort.
		 */
		private void mergeForceCollapse() {
			while (stackSize > 1) {
				int n = stackSize - 2;
				if (n > 0 && runLen[n - 1] < runLen[n + 1])
					n--;
				mergeAt(n);
			}
		}

		/**
		 * Merges the two runs at stack indices i and i+1. Run i must be the
		 * penultimate or antepenultimate run on the stack. In other words, i
		 * must be equal to stackSize-2 or stackSize-3.
		 * 
		 * @param i
		 *            stack index of the first of the two runs to merge
		 */
		private void mergeAt(int i) {
			assert stackSize >= 2;
			assert i >= 0;
			assert i == stackSize - 2 || i == stackSize - 3;

			int base1 = runBase[i];
			int len1 = runLen[i];
			int base2 = runBase[i + 1];
			int len2 = runLen[i + 1];
			assert len1 > 0 && len2 > 0;
			assert base1 + len1 == base2;

			/*
			 * Record the length of the combined runs; if i is the 3rd-last run
			 * now, also slide over the last run (which isn't involved in this
			 * merge). The current run (i+1) goes away in any case.
			 */
			runLen[i] = len1 + len2;
			if (i == stackSize - 3) {
				runBase[i + 1] = runBase[i + 2];
				runLen[i + 1] = runLen[i + 2];
			}
			stackSize--;

			K key0 = s.newKey();

			/*
			 * Find where the first element of run2 goes in run1. Prior elements
			 * in run1 can be ignored (because they're already in place).
			 */
			int k = gallopRight(s.getKey(a, base2, key0), a, base1, len1, 0, c);
			assert k >= 0;
			base1 += k;
			len1 -= k;
			if (len1 == 0)
				return;

			/*
			 * Find where the last element of run1 goes in run2. Subsequent
			 * elements in run2 can be ignored (because they're already in
			 * place).
			 */
			len2 = gallopLeft(s.getKey(a, base1 + len1 - 1, key0), a, base2, len2, len2 - 1, c);
			assert len2 >= 0;
			if (len2 == 0)
				return;

			// Merge remaining runs, using tmp array with min(len1, len2)
			// elements
			if (len1 <= len2)
				mergeLo(base1, len1, base2, len2);
			else
				mergeHi(base1, len1, base2, len2);
		}

		/**
		 * Locates the position at which to insert the specified key into the
		 * specified sorted range; if the range contains an element equal to
		 * key, returns the index of the leftmost equal element.
		 * 
		 * @param key
		 *            the key whose insertion point to search for
		 * @param a
		 *            the array in which to search
		 * @param base
		 *            the index of the first element in the range
		 * @param len
		 *            the length of the range; must be > 0
		 * @param hint
		 *            the index at which to begin the search, 0 <= hint < n. The
		 *            closer hint is to the result, the faster this method will
		 *            run.
		 * @param c
		 *            the comparator used to order the range, and to search
		 * @return the int k, 0 <= k <= n such that a[b + k - 1] < key <= a[b +
		 *         k], pretending that a[b - 1] is minus infinity and a[b + n]
		 *         is infinity. In other words, key belongs at index b + k; or
		 *         in other words, the first k elements of a should precede key,
		 *         and the last n - k should follow it.
		 */
		private int gallopLeft(K key, Buffer a, int base, int len, int hint, Comparator<? super K> c) {
			assert len > 0 && hint >= 0 && hint < len;
			int lastOfs = 0;
			int ofs = 1;
			K key0 = s.newKey();

			if (c.compare(key, s.getKey(a, base + hint, key0)) > 0) {
				// Gallop right until a[base+hint+lastOfs] < key <=
				// a[base+hint+ofs]
				int maxOfs = len - hint;
				while (ofs < maxOfs && c.compare(key, s.getKey(a, base + hint + ofs, key0)) > 0) {
					lastOfs = ofs;
					ofs = (ofs << 1) + 1;
					if (ofs <= 0) // int overflow
						ofs = maxOfs;
				}
				if (ofs > maxOfs)
					ofs = maxOfs;

				// Make offsets relative to base
				lastOfs += hint;
				ofs += hint;
			} else { // key <= a[base + hint]
				// Gallop left until a[base+hint-ofs] < key <=
				// a[base+hint-lastOfs]
				final int maxOfs = hint + 1;
				while (ofs < maxOfs && c.compare(key, s.getKey(a, base + hint - ofs, key0)) <= 0) {
					lastOfs = ofs;
					ofs = (ofs << 1) + 1;
					if (ofs <= 0) // int overflow
						ofs = maxOfs;
				}
				if (ofs > maxOfs)
					ofs = maxOfs;

				// Make offsets relative to base
				int tmp = lastOfs;
				lastOfs = hint - ofs;
				ofs = hint - tmp;
			}
			assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;

			/*
			 * Now a[base+lastOfs] < key <= a[base+ofs], so key belongs
			 * somewhere to the right of lastOfs but no farther right than ofs.
			 * Do a binary search, with invariant a[base + lastOfs - 1] < key <=
			 * a[base + ofs].
			 */
			lastOfs++;
			while (lastOfs < ofs) {
				int m = lastOfs + ((ofs - lastOfs) >>> 1);

				if (c.compare(key, s.getKey(a, base + m, key0)) > 0)
					lastOfs = m + 1; // a[base + m] < key
				else
					ofs = m; // key <= a[base + m]
			}
			assert lastOfs == ofs; // so a[base + ofs - 1] < key <= a[base +
			// ofs]
			return ofs;
		}

		/**
		 * Like gallopLeft, except that if the range contains an element equal
		 * to key, gallopRight returns the index after the rightmost equal
		 * element.
		 * 
		 * @param key
		 *            the key whose insertion point to search for
		 * @param a
		 *            the array in which to search
		 * @param base
		 *            the index of the first element in the range
		 * @param len
		 *            the length of the range; must be > 0
		 * @param hint
		 *            the index at which to begin the search, 0 <= hint < n. The
		 *            closer hint is to the result, the faster this method will
		 *            run.
		 * @param c
		 *            the comparator used to order the range, and to search
		 * @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b +
		 *         k]
		 */
		private int gallopRight(K key, Buffer a, int base, int len, int hint, Comparator<? super K> c) {
			assert len > 0 && hint >= 0 && hint < len;

			int ofs = 1;
			int lastOfs = 0;
			K key1 = s.newKey();

			if (c.compare(key, s.getKey(a, base + hint, key1)) < 0) {
				// Gallop left until a[b+hint - ofs] <= key < a[b+hint -
				// lastOfs]
				int maxOfs = hint + 1;
				while (ofs < maxOfs && c.compare(key, s.getKey(a, base + hint - ofs, key1)) < 0) {
					lastOfs = ofs;
					ofs = (ofs << 1) + 1;
					if (ofs <= 0) // int overflow
						ofs = maxOfs;
				}
				if (ofs > maxOfs)
					ofs = maxOfs;

				// Make offsets relative to b
				int tmp = lastOfs;
				lastOfs = hint - ofs;
				ofs = hint - tmp;
			} else { // a[b + hint] <= key
				// Gallop right until a[b+hint + lastOfs] <= key < a[b+hint +
				// ofs]
				int maxOfs = len - hint;
				while (ofs < maxOfs && c.compare(key, s.getKey(a, base + hint + ofs, key1)) >= 0) {
					lastOfs = ofs;
					ofs = (ofs << 1) + 1;
					if (ofs <= 0) // int overflow
						ofs = maxOfs;
				}
				if (ofs > maxOfs)
					ofs = maxOfs;

				// Make offsets relative to b
				lastOfs += hint;
				ofs += hint;
			}
			assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;

			/*
			 * Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere
			 * to the right of lastOfs but no farther right than ofs. Do a
			 * binary search, with invariant a[b + lastOfs - 1] <= key < a[b +
			 * ofs].
			 */
			lastOfs++;
			while (lastOfs < ofs) {
				int m = lastOfs + ((ofs - lastOfs) >>> 1);

				if (c.compare(key, s.getKey(a, base + m, key1)) < 0)
					ofs = m; // key < a[b + m]
				else
					lastOfs = m + 1; // a[b + m] <= key
			}
			assert lastOfs == ofs; // so a[b + ofs - 1] <= key < a[b + ofs]
			return ofs;
		}

		/**
		 * Merges two adjacent runs in place, in a stable fashion. The first
		 * element of the first run must be greater than the first element of
		 * the second run (a[base1] > a[base2]), and the last element of the
		 * first run (a[base1 + len1-1]) must be greater than all elements of
		 * the second run.
		 * 
		 * For performance, this method should be called only when len1 <= len2;
		 * its twin, mergeHi should be called if len1 >= len2. (Either method
		 * may be called if len1 == len2.)
		 * 
		 * @param base1
		 *            index of first element in first run to be merged
		 * @param len1
		 *            length of first run to be merged (must be > 0)
		 * @param base2
		 *            index of first element in second run to be merged (must be
		 *            aBase + aLen)
		 * @param len2
		 *            length of second run to be merged (must be > 0)
		 */
		private void mergeLo(int base1, int len1, int base2, int len2) {
			assert len1 > 0 && len2 > 0 && base1 + len1 == base2;

			// Copy first run into temp array
			Buffer a = this.a; // For performance
			Buffer tmp = ensureCapacity(len1);
			s.copyRange(a, base1, tmp, 0, len1);

			int cursor1 = 0; // Indexes into tmp array
			int cursor2 = base2; // Indexes int a
			int dest = base1; // Indexes int a

			// Move first element of second run and deal with degenerate cases
			s.copyElement(a, cursor2++, a, dest++);
			if (--len2 == 0) {
				s.copyRange(tmp, cursor1, a, dest, len1);
				return;
			}
			if (len1 == 1) {
				s.copyRange(a, cursor2, a, dest, len2);
				s.copyElement(tmp, cursor1, a, dest + len2); // Last elt of run
				// 1 to end of
				// merge
				return;
			}

			K key0 = s.newKey();
			K key1 = s.newKey();

			Comparator<? super K> c = this.c; // Use local variable for
			// performance
			int minGallop = this.minGallop; // "    " "     " "
			outer: while (true) {
				int count1 = 0; // Number of times in a row that first run won
				int count2 = 0; // Number of times in a row that second run won

				/*
				 * Do the straightforward thing until (if ever) one run starts
				 * winning consistently.
				 */
				do {
					assert len1 > 1 && len2 > 0;
					if (c.compare(s.getKey(a, cursor2, key0), s.getKey(tmp, cursor1, key1)) < 0) {
						s.copyElement(a, cursor2++, a, dest++);
						count2++;
						count1 = 0;
						if (--len2 == 0)
							break outer;
					} else {
						s.copyElement(tmp, cursor1++, a, dest++);
						count1++;
						count2 = 0;
						if (--len1 == 1)
							break outer;
					}
				} while ((count1 | count2) < minGallop);

				/*
				 * One run is winning so consistently that galloping may be a
				 * huge win. So try that, and continue galloping until (if ever)
				 * neither run appears to be winning consistently anymore.
				 */
				do {
					assert len1 > 1 && len2 > 0;
					count1 = gallopRight(s.getKey(a, cursor2, key0), tmp, cursor1, len1, 0, c);
					if (count1 != 0) {
						s.copyRange(tmp, cursor1, a, dest, count1);
						dest += count1;
						cursor1 += count1;
						len1 -= count1;
						if (len1 <= 1) // len1 == 1 || len1 == 0
							break outer;
					}
					s.copyElement(a, cursor2++, a, dest++);
					if (--len2 == 0)
						break outer;

					count2 = gallopLeft(s.getKey(tmp, cursor1, key0), a, cursor2, len2, 0, c);
					if (count2 != 0) {
						s.copyRange(a, cursor2, a, dest, count2);
						dest += count2;
						cursor2 += count2;
						len2 -= count2;
						if (len2 == 0)
							break outer;
					}
					s.copyElement(tmp, cursor1++, a, dest++);
					if (--len1 == 1)
						break outer;
					minGallop--;
				} while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
				if (minGallop < 0)
					minGallop = 0;
				minGallop += 2; // Penalize for leaving gallop mode
			} // End of "outer" loop
			this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to
			// field

			if (len1 == 1) {
				assert len2 > 0;
				s.copyRange(a, cursor2, a, dest, len2);
				s.copyElement(tmp, cursor1, a, dest + len2); // Last elt of run
				// 1 to end of
				// merge
			} else if (len1 == 0) {
				throw new IllegalArgumentException("Comparison method violates its general contract!");
			} else {
				assert len2 == 0;
				assert len1 > 1;
				s.copyRange(tmp, cursor1, a, dest, len1);
			}
		}

		/**
		 * Like mergeLo, except that this method should be called only if len1
		 * >= len2; mergeLo should be called if len1 <= len2. (Either method may
		 * be called if len1 == len2.)
		 * 
		 * @param base1
		 *            index of first element in first run to be merged
		 * @param len1
		 *            length of first run to be merged (must be > 0)
		 * @param base2
		 *            index of first element in second run to be merged (must be
		 *            aBase + aLen)
		 * @param len2
		 *            length of second run to be merged (must be > 0)
		 */
		private void mergeHi(int base1, int len1, int base2, int len2) {
			assert len1 > 0 && len2 > 0 && base1 + len1 == base2;

			// Copy second run into temp array
			Buffer a = this.a; // For performance
			Buffer tmp = ensureCapacity(len2);
			s.copyRange(a, base2, tmp, 0, len2);

			int cursor1 = base1 + len1 - 1; // Indexes into a
			int cursor2 = len2 - 1; // Indexes into tmp array
			int dest = base2 + len2 - 1; // Indexes into a

			K key0 = s.newKey();
			K key1 = s.newKey();

			// Move last element of first run and deal with degenerate cases
			s.copyElement(a, cursor1--, a, dest--);
			if (--len1 == 0) {
				s.copyRange(tmp, 0, a, dest - (len2 - 1), len2);
				return;
			}
			if (len2 == 1) {
				dest -= len1;
				cursor1 -= len1;
				s.copyRange(a, cursor1 + 1, a, dest + 1, len1);
				s.copyElement(tmp, cursor2, a, dest);
				return;
			}

			Comparator<? super K> c = this.c; // Use local variable for
			// performance
			int minGallop = this.minGallop; // "    " "     " "
			outer: while (true) {
				int count1 = 0; // Number of times in a row that first run won
				int count2 = 0; // Number of times in a row that second run won

				/*
				 * Do the straightforward thing until (if ever) one run appears
				 * to win consistently.
				 */
				do {
					assert len1 > 0 && len2 > 1;
					if (c.compare(s.getKey(tmp, cursor2, key0), s.getKey(a, cursor1, key1)) < 0) {
						s.copyElement(a, cursor1--, a, dest--);
						count1++;
						count2 = 0;
						if (--len1 == 0)
							break outer;
					} else {
						s.copyElement(tmp, cursor2--, a, dest--);
						count2++;
						count1 = 0;
						if (--len2 == 1)
							break outer;
					}
				} while ((count1 | count2) < minGallop);

				/*
				 * One run is winning so consistently that galloping may be a
				 * huge win. So try that, and continue galloping until (if ever)
				 * neither run appears to be winning consistently anymore.
				 */
				do {
					assert len1 > 0 && len2 > 1;
					count1 = len1 - gallopRight(s.getKey(tmp, cursor2, key0), a, base1, len1, len1 - 1, c);
					if (count1 != 0) {
						dest -= count1;
						cursor1 -= count1;
						len1 -= count1;
						s.copyRange(a, cursor1 + 1, a, dest + 1, count1);
						if (len1 == 0)
							break outer;
					}
					s.copyElement(tmp, cursor2--, a, dest--);
					if (--len2 == 1)
						break outer;

					count2 = len2 - gallopLeft(s.getKey(a, cursor1, key0), tmp, 0, len2, len2 - 1, c);
					if (count2 != 0) {
						dest -= count2;
						cursor2 -= count2;
						len2 -= count2;
						s.copyRange(tmp, cursor2 + 1, a, dest + 1, count2);
						if (len2 <= 1) // len2 == 1 || len2 == 0
							break outer;
					}
					s.copyElement(a, cursor1--, a, dest--);
					if (--len1 == 0)
						break outer;
					minGallop--;
				} while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
				if (minGallop < 0)
					minGallop = 0;
				minGallop += 2; // Penalize for leaving gallop mode
			} // End of "outer" loop
			this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to
			// field

			if (len2 == 1) {
				assert len1 > 0;
				dest -= len1;
				cursor1 -= len1;
				s.copyRange(a, cursor1 + 1, a, dest + 1, len1);
				s.copyElement(tmp, cursor2, a, dest); // Move first elt of run2
				// to front of merge
			} else if (len2 == 0) {
				throw new IllegalArgumentException("Comparison method violates its general contract!");
			} else {
				assert len1 == 0;
				assert len2 > 0;
				s.copyRange(tmp, 0, a, dest - (len2 - 1), len2);
			}
		}

		/**
		 * Ensures that the external array tmp has at least the specified number
		 * of elements, increasing its size if necessary. The size increases
		 * exponentially to ensure amortized linear time complexity.
		 * 
		 * @param minCapacity
		 *            the minimum required capacity of the tmp array
		 * @return tmp, whether or not it grew
		 */
		private Buffer ensureCapacity(int minCapacity) {
			if (tmpLength < minCapacity) {
				// Compute smallest power of 2 > minCapacity
				int newSize = minCapacity;
				newSize |= newSize >> 1;
				newSize |= newSize >> 2;
				newSize |= newSize >> 4;
				newSize |= newSize >> 8;
				newSize |= newSize >> 16;
				newSize++;

				if (newSize < 0) // Not bloody likely!
					newSize = minCapacity;
				else
					newSize = Math.min(newSize, aLength >>> 1);

				tmp = s.allocate(newSize);
				tmpLength = newSize;
			}
			return tmp;
		}
	}
}
